French Tarot is played with a 78-card deck.
The deck includes four suits (hearts, diamonds, spades, clubs).
Each suit contains 14 cards: 1, 2, ..., 10, Jack, Knight, Queen, King.
There are also 21 numbered trump cards and one unnumbered Fool trump card.
Four players are dealt 18-card hands and the dog is dealt 6 cards.
Knowing that I have 5 hearts in my hand what is the probability that the other player with the most hearts has 4 hearts?
I tried:
$\dfrac{ \binom{9}{4}\cdot\binom{69}{14}}{\binom{78}{18} }=0.0916$
but I know the correct result is 0.48 (from a Tarot book which includes probability tables).
I think I'm failing to have the proper conditional probability for the 3 other players and the dog. Any help would be much appreciated.
Once you're dealt your hand, which is stipulated to have $5$ hearts and $13$ non-hearts, the remaining $60$-card deck consists of $9$ hearts and $51$ non-hearts. Now, let $P_{a,b,c}$ be the probability that players $A$, $B$, and $C$ are dealt exactly $a$, $b$, and $c$ hearts from that remaining deck, respectively, where $3 \le a+b+c\le 9$: $$ P_{a,b,c}=\frac{\frac{9!}{a! b! c!(9-a-b-c)!}\cdot\frac{51!}{ (18-a)! (18-b)! (18-c)! (a+b+c-3)!}}{\frac{60!}{18! 18! 18! 6!}}. $$ Here we've counted the ways to choose which hearts each player receives (subject to the $abc$-constraint), times the ways to choose which non-hearts each player receives (ditto), divided by the ways to choose the $18$-card hands independent of the $abc$-constraint.
You want the probability that one of $\{A,B,C\}$ has $4$ hearts, and none of them has more than $4$. That leaves these possibilities for $(a,b,c)$, up to permutations: $(4,0,0)$, $(4,1,0)$, $(4,1,1)$, $(4,2,0)$, $(4,2,1)$, $(4,2,2)$, $(4,3,0)$, $(4,3,1)$, $(4,3,2)$, $(4,4,0)$, $(4,4,1)$. The result is $$ \begin{eqnarray} p_4 &=& 3P_{4,0,0}+6P_{4,1,0} + 3P_{4,1,1}+6P_{4,2,0}+6P_{4,2,1} \\ &&+3P_{4,2,2}+6P_{4,3,0}+6P_{4,3,1}+6P_{4,3,2}+3P_{4,4,0}+3P_{4,4,1} \\ &\approx& 0.4720001, \end{eqnarray} $$ where the coefficients account for the number of distinct permutations of $(a,b,c)$. Similarly, to cover the remaining options: the probability that the player besides you with the most hearts has fewer than $4$ hearts is $p_{<4}\approx 0.2841916$, and the probability that he has more than $4$ hearts is $p_{>4}\approx 0.2438084$. The three options add up to $1$, as they must.